# Proving that Watt's linkage traces the lemniscate of Bernoulli

I'm trying to prove that, as Wikipedia says, "when the lengths of its bars and its base are chosen to form a crossed square, it traces the lemniscate of Bernoulli" I'm using the setup described here (Although, they have a mistake the fixed point in the text should be D, as in the diagram), but all multiplied by a factor of 2.

So I decided to use vectors to prove it.

Let the point B and C have position vectors $\vec{x}$ and $\vec{x}'$ respectivelly. And the point (1, 0) be $\vec{a}$ . Then the constraints are: $$\begin{gather} (\vec{x}'-\vec{a}) \cdot (\vec{x}'-\vec{a}) =2 \\ (\vec{x}+\vec{a}) \cdot (\vec{x}+\vec{a}) =2.\end{gather}$$

Also, if we let $\vec{X}$ be the position vector to trace the lemniscate, then: $$\vec{X}=\vec{x}+1/2(\vec{x}'-\vec{x})=1/2(\vec{x}'+\vec{x})=1/2(\vec{x}'-\vec{a}+\vec{x}+\vec{a}).$$

Now, \begin{align} (\vec{X} \cdot \vec{X})^2 &=(1/4((\vec{x}'-\vec{a}) \cdot (\vec{x}'-\vec{a})+(\vec{x}+\vec{a}) \cdot (\vec{x}+\vec{a})+2(\vec{x}'-\vec{a}) \cdot (\vec{x}+\vec{a})))^2 \\ &=(1/4(4+2(\vec{x} \cdot \vec{x}'-\vec{a} \cdot \vec{x} +\vec{a} \cdot \vec{x}' -\vec{a} \cdot \vec{a}))^2 \\ &=(1/2(1+xx'+yy'-x+x'))^2\end{align}.

From which I just can't get the required $2(X^2-Y^2)=1/2((x+x')^2-(y+y')^2)$ on the RHS to make the algebraic equation for the lemniscate. In particular, I can't get any $y^2$ or $y'^2$. So what am I doing wrong?